Question

a ball is dropped from $405$ meters and rebounds two-thirds the distance it falls each time it bounces. how many meters will the ball have traveled when it hits the ground the fourth time?

Answer 1

The total distance traveled by the ball when it hits the ground for the fourth time is the sum of the individual distances fallen and bounced, starting from 405 meters and subsequently being two-thirds of the previous distance for each bounce.

Step-by-step explanation:

The problem involves a ball dropped from a height and bouncing back to a fraction of the height it fell from. To calculate the total distance traveled by the ball when it hits the ground for the fourth time, we'll sum the distances of all the falls and bounces until the fourth hit. The initial fall is from 405 meters, and each subsequent bounce reaches two-thirds of the previous height.

1. First fall = 405 meters.
2. First bounce up = 2/3 of first fall = (2/3) * 405 meters.
3. Second fall = First bounce up.
4. Second bounce up = 2/3 of second fall = (2/3)^2 * 405 meters.
5. Third fall = Second bounce up.
6. Third bounce up = 2/3 of third fall = (2/3)^3 * 405 meters.
7. Fourth fall = Third bounce up.

The total distance when the ball hits the ground for the fourth time will be the sum of these seven distances:

Total distance = First fall + First bounce up + Second fall + Second bounce up + Third fall + Third bounce up + Fourth fall.

By plugging in the values, we can calculate the exact distance.